A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine

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Solution

pH = 3.44
We know that,
$p H = - l o g [H^{+}] ∴ [H^{+}] = 3.63 \times 10^{- 4}$
Then, $K_{b} = \frac{(3.633.63 \times 10^{- 4})^{2}}{0.02}$ ( $∵$ concentration =0.02M)
$\Rightarrow K_{b} = 6.6 \times 10^{- 6}$
Now, $K_{b} = \frac{K_{w}}{K_{a}}$
$\Rightarrow K_{a} = \frac{K_{w}}{K_{a}} = \frac{10^{- 14}}{6.6 \times 10^{- 6}}$
$= 1.51 \times 10^{- 9}$

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A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine

pH = 3.44 We know that, pH=–log[H+]∴[H+]=3.63×10−4 Then, Kb=(3.633.63×10−4)20.02 (∵ concentration =0.02M) ⇒Kb=6.6×10−6 Now, Kb=KwKa ⇒Ka=KwKa=10−146.6×10−6 =1.51×10−9